I am kind of confused in reading the next proof of a theorem in "Stochastic Analysis: Itô and Malliavin Calculus" in Tandem by Matsumoto & Taniguchi;
Here, a closable submartingale $\{ Z_n \}_n$ is defined as a submartingale that is bounded by an integrable random variable $Z$ in $$ Z_n \leq \textrm{E}[Z|\mathcal{F}_n]. $$
Where I am stuck: The equation in the middle, \begin{equation} \mathrm{E}[Y|\mathcal{F}_n] = \lim_{m \rightarrow \infty} \mathrm{E}[X_m^+ | \mathcal{F}_n]. \tag{1} \label{eq:question} \end{equation}
I don't see why we can get the limit out of the expectation. The text says "since $X_n^+$ converges also in $L^1$", but I don't think when $X_n \rightarrow X$ in $L^1$, it always holds for a sub-$\sigma$-algebra $\mathcal{G}$, $$ \textrm{E}[X_n | \mathcal{G}] \rightarrow \textrm{E}[X | \mathcal{G}] \quad \textrm{a.s.} $$ For we have a counterexample: we take $\mathcal{G} = \mathcal{F}$ (where $\mathcal{F}$ is the $\sigma$-algebra of the whole probability space) and $\{X_n\}$ such that $X_n \rightarrow X$ in $L^1$ but not $\textrm{a.s.}$, then $$ \textrm{E}[X_n | \mathcal{F}] = X_n \not\rightarrow X = \textrm{E}[X | \mathcal{F}] \quad \textrm{a.s.} $$ So how can we validate Eq.\eqref{eq:question}?
$E(Y|\mathcal F_n) \geq X_n^{+}$ $P-$ as. is correct but the first equality may not hold. Use the fact that $L^{1}$ convergence implies a.s. convergence for a subsequece Since $E(X_m^{+}|\mathcal F_n)\to E(Y|\mathcal F_n)$ in $L^{1}$ as $m \to \infty$we can go to a subsequence to finish.